Test Index

Quadratic Equations

Section: Mathematics

Question : 1 of 51

Marks: +1, -0

\sqrt{3 x^2+x+5}=x-3

Solution:

Given, equation is

\sqrt{3 x^2+x+5}=x-3

Here,

3 x^2+x+5 \geq 0

(since square root is non-negative)

\Rightarrow x-3 \geq 0

\Rightarrow x \geq 3

Now, squaring to both sides of given equation, we get

\left(\sqrt{3 x^2+x+5}\right)^2=(x-3)^2

\Rightarrow 3 x^2+x+5=x^2-6 x+9

\Rightarrow 2 x^2+7 x-4=0

Using quadratic formula,

x=\frac{-7 \pm \sqrt{7^2-4 \cdot 2 \cdot (-4)}}{2 \cdot 2}

= \frac{-7 \pm \sqrt{49+32}}{4} = \frac{-7 \pm 9}{4}

\Rightarrow x=\frac{2}{4} \text{ and } x=\frac{-16}{4}

\Rightarrow x=\frac{1}{2} \text{ and } x=-4

Since domain,

x \geq 3

, so

x=-4

And

x=\frac{1}{2}

are not possible.

∴ There is zero solution.

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